Find the reference angle.

Here are the solutions to the trigonometric equations.

: Find the value of $\theta$ between $0^\circ$ and $360^\circ$ for the equation $\sin(2\theta) = 0.5$.

Step 1: Find the reference angle.
Let $X = 2\theta$. The equation becomes $\sin(X) = 0.5$.
The reference angle $\alpha$ for $\sin(\alpha) = 0.5$ is:

$$\alpha = \sin^{-1}(0.5) = 30^\circ$$

Step 2: Determine the quadrants where $\sin(X)$ is positive.
Since $\sin(X)$ is positive, $X$ must be in the 1st or 2nd quadrant.

Step 3: Find all possible values for $X$ in the range $0^\circ \le X \le 720^\circ$.
Since $0^\circ \le \theta \le 360^\circ$, the range for $X = 2\theta$ is $0^\circ \le 2\theta \le 720^\circ$.
In the 1st quadrant:

$$X = 30^\circ$$ $$X = 30^\circ + 360^\circ = 390^\circ$$

In the 2nd quadrant:

$$X = 180^\circ - 30^\circ = 150^\circ$$ $$X = 150^\circ + 360^\circ = 510^\circ$$

So, the values for $X$ are $30^\circ, 150^\circ, 390^\circ, 510^\circ$.

Step 4: Solve for $\theta$.
Substitute $X = 2\theta$ and divide by 2:

$$2\theta = 30^\circ \implies \theta = \frac{30^\circ}{2} = 15^\circ$$ $$2\theta = 150^\circ \implies \theta = \frac{150^\circ}{2} = 75^\circ$$ $$2\theta = 390^\circ \implies \theta = \frac{390^\circ}{2} = 195^\circ$$ $$2\theta = 510^\circ \implies \theta = \frac{510^\circ}{2} = 255^\circ$$

All these values are within the range $0^\circ \le \theta \le 360^\circ$.

The values of $\theta$ are $\boxed{15^\circ, 75^\circ, 195^\circ, 255^\circ}$.

: Find the value of $\theta$ to the nearest degree for $\cos(3\theta + 20^\circ) = \frac{\sqrt{3}}{2}$ where $0^\circ \le \theta \le 360^\circ$.

Step 1: Find the reference angle.
Let $Y = 3\theta + 20^\circ$. The equation becomes $\cos(Y) = \frac{\sqrt{3}}{2}$.
The reference angle $\alpha$ for $\cos(\alpha) = \frac{\sqrt{3}}{2}$ is:

$$\alpha = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = 30^\circ$$

Step 2: Determine the quadrants where $\cos(Y)$ is positive.
Since $\cos(Y)$ is positive, $Y$ must be in the 1st or 4th quadrant.

Step 3: Determine the range for $Y$.
Given $0^\circ \le \theta \le 360^\circ$:
Multiply by 3: $0^\circ \le 3\theta \le 1080^\circ$
Add $20^\circ$: $0^\circ + 20^\circ \le 3\theta + 20^\circ \le 1080^\circ + 20^\circ$
So, $20^\circ \le Y \le 1100^\circ$.

Step 4: Find all possible values for $Y$ in the range $20^\circ \le Y \le 1100^\circ$.
In the 1st quadrant:

$$Y = 30^\circ$$ $$Y = 30^\circ + 360^\circ = 390^\circ$$ $$Y = 30^\circ + 720^\circ = 750^\circ$$

In the 4th quadrant:

$$Y = 360^\circ - 30^\circ = 330^\circ$$ $$Y = 330^\circ + 360^\circ = 690^\circ$$ $$Y = 330^\circ + 720^\circ = 1050^\circ$$

The next values ($30^\circ + 1080^\circ = 1110^\circ$ and $330^\circ + 1080^\circ = 1410^\circ$) are outside the range $Y \le 1100^\circ$.
So, the values for $Y$ are $30^\circ, 330^\circ, 390^\circ, 690^\circ, 750^\circ, 1050^\circ$.

Step 5: Solve for $\theta$ and round to the nearest degree.
Substitute $Y = 3\theta + 20^\circ$:

$$3\theta + 20^\circ = 30^\circ \implies 3\theta = 10^\circ \implies \theta = \frac{10^\circ}{3} \approx 3.33^\circ \approx 3^\circ$$ $$3\theta + 20^\circ = 330^\circ \implies 3\theta = 310^\circ \implies \theta = \frac{310^\circ}{3} \approx 103.33^\circ \approx 103^\circ$$ $$3\theta + 20^\circ = 390^\circ \implies 3\theta = 370^\circ \implies \theta = \frac{370^\circ}{3} \approx 123.33^\circ \approx 123^\circ$$ $$3\theta + 20^\circ = 690^\circ \implies 3\theta = 670^\circ \implies \theta = \frac{670^\circ}{3} \approx 223.33^\circ \approx 223^\circ$$ $$3\theta + 20^\circ = 750^\circ \implies 3\theta = 730^\circ \implies \theta = \frac{730^\circ}{3} \approx 243.33^\circ \approx 243^\circ$$ $$$$